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Okay, so on a Calculus AB test on telling extrema, concavity, etc. by the 1st and 2nd derivative of a function, we were given a problem as follows.

We were given a graph, labeled $y=p'(x)$. It was piece-wise, but the only important part for this question looks like this:

derivative graph

We were told that there was a vertical tangent at $x=4$. The curve passes through $(4,0)$ and it is clear that when $x<4$, $p'(x)>0$ and when $x>4$, $p'(x)<0$.

Anyways, the question was, where is $p(x)$ concave down. So the way to determine that would be via $p''(x)$, or by just checking where $p'(x)$ is decreasing.

This is what I put:

(something1, 4)U(4, something2]

But according to my teacher, the answer is:

(something1, something2]

He is cool and very open-minded about it and is going to check up on it, and tell me tomorrow. But I'm impatient.

My reasoning is that, because $p''(x)$ approaches $-\infty$ as $x$ approaches $4$, the concavity would be undefined. If we graphed $p(x)$, it would clearly look concave down, but I am convinced that it isn't concave down right at $x=4$.

I figure that all depends on how we define an undefined second derivative. So that's my question. What is the concavity at a point if the second derivative at that point is undefined? My guess would be that the concavity would be undefined.

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3 Answers 3

I would agree with your teacher. The function is concave down on $(-\infty,\mbox{something bigger than 4}]$.

Yes. The second derivative is undefined at $x=4$, but this doesn't negate the possibility of being concave down. The function is concave down if the derivative is decreasing. Agree? Well, looking at your derivative it's decreasing even at $x=4$.

This is related to the fact that a if a function is increasing, we are not guaranteed that the derivative exists (and is positive).

Checking the second derivative is a test for concavity. If the second derivative does not exist, the test does not apply.

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Thanks for the reply. But if a tangent line is vertical, can we really say that the function is decreasing there? If you look at the function around it, you would clearly say, yes, it is decreasing; but right AT that point, can we say that? I think I am just confused maybe on how a slope of infinity and a slope of negative infinity appear the same. (Without looking at the surrounding function). –  Brandon oubiub Oct 13 '11 at 23:39
    
Wait, I think I get it more. Because infinity slope does not equal negative infinity slope it just appears to. I'm treating infinity too much like a solid number I think. It is still decreasing, even though it is decreasing by a LOT. It is vertical, but it is vertical down. (Is it safe to say that it merely approaches being vertical?) –  Brandon oubiub Oct 13 '11 at 23:48
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Decreasing "at x=a" is a little fishy in many ways. We should really only talk about a function decreasing on an interval of $x$'s. If you check the original function, take $4-\epsilon < x < y < 4+\epsilon$ then you would see $f(x)>f(y)$ so the function is decreasing everywhere "near" $x=4$. Thus it decreases "at $x=4$". –  Bill Cook Oct 14 '11 at 0:20

The question your asking is an issue that a number of Calculus teachers were discussing at a conference I was at a few years ago. Many people were also curious about which notation to use for intervals of increasing, decreasing, and concavity. Do you use parentheses or brackets?

The answer is of course that you cannot be increasing, decreasing, concave up, or concave down at a single point. So by asking is it concave down at 4 is like asking if the water level in a picture of a pool is rising or falling. You are only looking a one point in time and it is not feasible to say either. Your teacher's interval is correct because we do not need to make any statements about the point where $x=4$.

If it makes you happy though, you could write it as:

Concave down on $(-\infty, 4) \cup (4, \text{something})$

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I think the if and only if definition of concave functions using negative $f''$ only apply if $f$ is twice-differentiable, which it is not.

The definition of a concave function doesn't actually have to rely on the second-derivatives though. In convex optimization people talk about convexity and concavity all the time on functions that are not even differentiable once.

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